Đáp án đúng: A
Giải chi tiết:\(\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {4{x^2} + x}  - \sqrt[3]{{8{x^3} - x}}} \right)\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {4{x^2} + x}  - 2x + 2x - \sqrt[3]{{8{x^3} - x}}} \right)\\ = \mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {4{x^2} + x}  - 2x} \right) + \mathop {\lim }\limits_{x \to  + \infty } \left( {2x - \sqrt[3]{{8{x^3} - x}}} \right)\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{4{x^2} + x - 4{x^2}}}{{\sqrt {4{x^2} + x}  + 2x}} + \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{8{x^3} - 8{x^3} + x}}{{4{x^2} + 2x.\sqrt[3]{{8{x^3} - x}} + {{\sqrt[3]{{8{x^3} - x}}}^2}}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{x}{{\sqrt {4{x^2} + x}  + 2x}} + \mathop {\lim }\limits_{x \to  + \infty } \dfrac{x}{{4{x^2} + 2x.\sqrt[3]{{8{x^3} - x}} + {{\sqrt[3]{{8{x^3} - x}}}^2}}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{1}{{\sqrt {4 + \dfrac{1}{x}}  + 2}} + \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{\dfrac{1}{x}}}{{4 + 2.\sqrt[3]{{8 - \dfrac{1}{{{x^2}}}}} + {{\sqrt[3]{{8 - \dfrac{1}{{{x^2}}}}}}^2}}}\\ = \dfrac{1}{{2 + 2}} + 0 = \dfrac{1}{4}\end{array}\)